Q107 / 200 · NEET 2005 · Physics

A stone tied to the end of a string of 1 $\mathrm{m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

  • Correct answer
    A
    $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre
  • B
    $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius away from the centre
  • C
    $\pi^2 \mathrm{~ms}^{-2}$ and direction along the tangent to the circle
  • D
    $\pi^2 / 4 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

Solution

\(a=\frac{v^2}{r}=\omega^2 r=4 \pi^2 n^2 r=4 \pi^2\left(\frac{22}{44}\right)^2 \times 1=\pi^2 \mathrm{~m} / \mathrm{s}^2\) and its direction is always along the radius and towards the centre.

Hence, the correct answer is option: 'A'.

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